A 10.0-g marble slides to the left with a velocity of magnitude 0.450 m/s on the
ID: 2259527 • Letter: A
Question
A 10.0-g marble slides to the left with a velocity of magnitude 0.450 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 20.0-g marble sliding to the right with a velocity of magnitude 0.250 m/s.
A 10.0-g marble slides to the left with a velocity of magnitude 0.450 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 20.0-g marble sliding to the right with a velocity of magnitude 0.250 m/s. Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. Take right as the positive x direction.) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble. Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble.Explanation / Answer
A)Let new velocities be Vsmall and Vlarge , then by conservation of momemtum,
-10X0.45 + 20 X 0.250 = 10 X Vsmall + 20 X Vlarge
so , Vsmall +(2 X Vlarge) = 0.05 or
Vsmall = 0.05- 2X Vlarge
Also by conservation of energy ,
[(1/2) X 10 X (.45)^2] + [(1/2) X 20 X (.25)^2]= [(1/2) X 10 X (Vsmall)^2] + [(1/2) X 20 X (Vlarge)^2]
solving,
.3275= (Vsmall)^2 + 2X (Vlarge)^2
But ,
Vsmall = 0.05- 2X Vlarge , so
.3275= ( 0.05- 2X Vlarge)^2 + 2X (Vlarge)^2
so , Vlarge= -0.217 m/s , Vsmall = 0.484 m/s
quadratic will give 2 roots , choose sign by knowing large marble will now bounce to opposite side that is left - negative(sign by convention)
B. Change in momemtum
Large Marble : Final - initial
= 20 /1000 X ( -.217 - 0.250) (mass of large marble = 20/1000 kg)
= -.00934 kgm/s (a decrease)
similarily for small marble
Change in momemtum = 10/1000 X ( .484 - (-.450))
= .00934 kgm/s a increase
C. Chane in Kinetic energy
Large Marble:
Initial Kinetic energy = 1/2 X (20/1000) x .250X.250= 6.25 X 10^(-4)
Final Kinetic energy = 1/2 X (20/1000) x .217X.217= 4.71 X 10^(-4)
So change = Decrease of 1.54 joules
small marble
By conservation of energy , small marble will gain equal amount= 1.54 J
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